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a ball is kicked at an angle of 40 degrees with the ground with the initial velocity of 15m/s. the ball reaches point b after 1.5 seconds. 1. what is the initial velocity in the y-direction? 2. what is the final velocity in the x-direction (at point b)? 3. what is the range, xb, in meters? 4. what is the height, hb, when the ball reaches point b, in meters?



Answer :

onyebuchinnaji

(1) The initial velocity in the y-direction is 9.64 m/s.

(2) The final velocity in the x-direction is 11.5 m/s.

(3)  The range, xb, of the ball is 22.6 m.

(4) The maximum height reached by the ball is 4.74 m.

What is the initial vertical component of the velocity?

The initial vertical component of the velocity is calculated using the following kinematic equation.

Vy = V sinθ

where;

  • V is the initial velocity
  • θ is the angle of projection of the ball

Vy = V sinθ

Vy = ( 15 m/s ) x ( sin 40 )

Vy = 9.64 m/s

The final velocity in the x-direction is calculated as follows;

Vₓf = Vxi = V cosθ

where;

  • Vₓf is the final horizontal velocity
  • Vₓi is the initial horizontal velocity

Vₓf  = ( 15 m/s ) x ( cos 40 )

Vₓf  = 11.5 m/s

The range of the ball is calculated as follows;

R = v² sin (2θ) / g

R = ( 15² sin (2 x 40) ) / 9.8

R = 22.6 m

The maximum height reached by the ball is calculated as;

H = v² sin²θ / 2g

H = ( 15² x ( sin 40 )² ) / (2 x 9.8)

H = 4.74 m

Learn more about maximum height of projectile here: https://brainly.com/question/12870645

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